Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(a, f(b, x))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, x)
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(a, f(b, x)))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, x))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(a, f(b, x))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, x)
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(a, f(b, x)))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, x))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(a, f(b, x))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(a, f(b, x)))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, x)
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, x))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.